# Metric on $M[0,1]$ the space of measures

Let $$B$$ be the closed unit ball on the space $$M[0,1]$$ of the borel regular complex measures on $$[0,1]$$. For $$\mu$$ and $$\nu$$ $$\in$$ $$M[0,1]$$ define $$d(u,v) = \sum_{n=0}^{\infty} \left|\int_{[0,1]}x^nd\mu - \int_{[0,1]}x^nd\nu \right|$$. Show that $$d$$ is a metric on $$M$$ and that define the weak star topology on $$B$$ but not in $$M$$.

I know that $$C_0([0,1])$$ is separable and hence I know that $$(K,w^*)$$ is metrizable with $$K$$ $$w^*$$-compact on $$M$$. And even more if $$(x_i)$$ is a dense set on the space then $$\sum_{n=1}^{\infty}\frac{p_n(\varphi - \psi)}{1 + p_n(\varphi - \psi)}$$ where $$p_n(\varphi) = \varphi(x_n)$$, is a metric which defines the $$w^*$$ topology on $$K$$. I would like to use this result but the polynomials given are not dense on the space. Another problem I have is that I am not able of see why the series given in the statement converges.

Another way I would like to attack this is proving that the identity function $$id:(K,w^*)\to (K,d)$$ is continuos, and since $$K$$ is $$w*$$ compact the result will follow.

I am stuck on this lines, I would appreciate any idea.

• Remember, Complex measures always have finite variation, by definition.

• As was mentioned in the comments, $$d$$ can not be a metric since it can take an infinite value, for example, when you put $$\nu = 0$$ and $$\mu$$ the lebesgue measure then you will get $$\int_{0}^1\frac{1}{1-x}$$ which diverges. A simpler example is considering dirac measures. So the problem

• I make a typo in the theorem I wanted to use, the metric associated with the dense subset $$(x_i)$$ is $$\sum_{n=1}^{\infty}\frac{1}{2^n}\frac{p_n(\varphi - \psi)}{1 + p_n(\varphi - \psi)}$$

Maybe it is worth mentioning that the idea of the theorem I wanted to use is to observe that if I truncate the series then I have a pseudometric which is $$w^*$$ continuos, and since the series converges uniformly, then the series is $$w^*$$continuos. Last, since $$(x_i)$$ is dense one can prove that actually is a metric, and since is continuos one already has that $$w^*$$ topology includes the topology induced by the metric, the final inclusion follows using that $$K$$ is $$w^*$$ compact.

EDIT2:

• I will write it to the author of the problem. But I think I have the essential key. I will expect that the correct $$d$$ on the problem is the one I give, if $$p_n(\mu) = \int_{0}^1x^n d\mu$$ then

$$d(\mu,\nu) = \sum_{n=1}^{\infty}\frac{1}{2^n}\frac{p_n(\mu - \nu)}{1 + p_n(\mu - \nu)}$$

in this case the truncated series is $$w^*$$ continuos, and since the series converges uniformly then $$d$$ is $$w^*$$ continuos, this implies that $$\tau_d \subset w^*$$. Now $$d$$ is a metric: if $$d(\mu,\nu) = 0$$ then each term of the sum is zero and then, by linearity, $$\int p(x)d\mu - \int p(x)d\nu = 0$$ with $$p(x)$$ being a polynomial, since the polynomials are dense on $$C_0([0,1])$$ and the integrals are linear functionals on $$C_0[0,1]$$ which are continuos, then this linear functionals are equal and then are equal the measures.

To finish, if we have that $$id:(K,w^*) \to (K,\tau_d)$$ is continuos, and since $$K^*$$ is $$w^*$$ compact then $$id$$ will be an homeomorphism.

I have not thought why this metric does not work on $$M$$

EDIT3:

Well finally, the author say he just forgot the factor $$\frac{1}{2^n}$$ in $$d$$, hence the metric is $$d(u,v) = \sum_{n=0}^{\infty}\frac{1}{2^n}\left|\int_0^1 x^n d\mu - \int_{0}^{1}x^n \right | d\nu$$

Whit this in mind I have a possible demonstration. I will propose it as answer to be discussed.

• Why is $x^n$ $\mu$-integrable? Do you mean finite Borel measures? Oct 6, 2019 at 3:28
• @amsmath, As far as I understand, complex measures automatically have finite total variation. But clarity would be always welcomed... Oct 6, 2019 at 3:34
• @SangchulLee Gosh, you're right. I forgot about that. Sorry. Oct 6, 2019 at 3:34
• This distance is not well defined. For example, take the Lebesgue measure as $\mu$ and $\nu = 0$. Then $$d(\mu,\nu) = \sum_n\int_0^1 x^n\,d\mu = \int_0^1\left(\sum_n x^n\right)\,dx = \int_0^1\frac 1{1-x}\,dx = \infty.$$ Oct 6, 2019 at 3:41
• I see no reason that $d$ is a metric on $B$, much less on $M[0,1]$. Indeed, $$\left| \int_{[0,1]}x^n \,\delta_1(\mathrm{d}x)-\int_{[0,1]}x^n \,\delta_0(\mathrm{d}x)\right|=\mathbf{1}_{n\geq1},$$ and so, its sum over all $n\geq0$ will diverge. Perhaps the author wanted to point out that the weak-$*$ topology on $B$ is the same as the uniform topology given by pseudometrics $$d_n(\mu,\nu)=\left|\int_{[0,1]}x^n\,\mu(\mathrm{d}x)-\int_{[0,1]}x^n\,\nu(\mathrm{d}x)\right|,$$ which in turn is the metric topology given by $$d(\mu,\nu)=\sum_{n=0}^{\infty} 2^{-n} d_n(\mu, \nu).$$ Oct 6, 2019 at 3:42

$$M[0,1]$$ cannot be $$w^*$$-metrizable because a dual space $$X^*$$ (of some Banach $$X$$) is $$w^*$$-metrizable if and only if $$X^*$$ is of finite dimension.

• Can you explain why? or give some bibliography?
– HFKy
Oct 6, 2019 at 15:09
• you may refer to math.stackexchange.com/questions/2486297/…
– Nick
Oct 6, 2019 at 15:20
• Okey, it seems a nice answer. The problem is that they use a lot of theorems that I am not suppose to know. However, reading the answers give me some inspiration. Instead of using the result your propose, what do you think of this:
– HFKy
Oct 6, 2019 at 15:36
• If it were metrizable, the open ball of center 0 and radius 1 will be open (in fact this is always true because $d$ is weak star continuos). Then take a point distinct of zero in this ball, let call it $T$. Since the ball is open there exists a local neighborhood of the weak star topology containing $T$ inside the open ball. However, since $X$ is of infinite dimension, this neighborhood will contain an infinite dimension subspace, in particular a line. Let $H$ be a generator of such line, then $T+\alpha H$ will be in $B(0,1)$ for all $\alpha$. Since $H$ is not zero and $d$ is a metric, then
– HFKy
Oct 6, 2019 at 15:41
• $H(x^n)$ is not zero for some $n$, and hence if $\alpha$ is greater enough then is absurd that $T+\alpha H \in B(0,1)$. Am I right?
– HFKy
Oct 6, 2019 at 15:42
• First, $$d(u,v) = \sum_{n=0}^{\infty}\frac{1}{2^n}\left|\int_0^1 x^n d\mu - \int_{0}^{1}x^n \right | d\nu$$ is metric on $$M[0,1]$$:
1. First the series is finite: this is because the total variation of the measures are finite and because $$x^n$$ is always less than 1, then this follows from triangular inequality.

2. The properties of a pseudometric are met: this is because the series is equal to $$d(u,v) = \sum_{n=0}^{\infty}\frac{1}{2^n} |p_n(\mu - \nu)|$$ and each $$p_n$$ are in fact seminorms.

3. Finally let's check that $$d(\mu,\nu) = 0$$ implies $$\mu = \nu$$. If this happens then the linear functionals associated $$T_\mu, T_\nu \in C_0([0,1])'$$ are equal on the polynomials of the form $$x^n$$, then by linearity are equal on the polynomials. Finally since the polynomials on $$[0,1]$$ are dense on $$C_0([0,1])$$ and the linear functionals continuous then they are equal on every function of $$C_0([0,1])$$, this implies $$\mu = \nu$$.

• $$d$$ gives the weak star topology on B, the closed unit ball with the norm of the total variation:
1. $$d$$ is continuos: $$d_k(\mu,\nu) = \sum_{n=0}^{k}\frac{1}{2^n}\left|\int_0^1 x^n d\mu - \int_{0}^{1}x^n \right | d\nu$$ is $$w^*$$-continuous, and it converges uniformly to $$d$$, this is because here the variaton of the measures is less than 1, and since $$x^n$$ is less than 1, the series defining $$d$$ is less than $$\sum_{n=0}^{\infty}\frac{1}{2^n}2 < \infty$$ and hence uniform convergence follows from Weierstrass M-Test. Uniform convergence implies $$w^*$$ continuity of $$d$$.

2. Since $$d$$ is continuos we have that the identity map $$id:(B,w^*) \to (B,\tau_d)$$ is continuous. Now since $$B$$ is $$w^*$$ compact, this identity map is a homeomorphism. This follows because if $$A$$ is a closed set in $$B$$ then is compact, then $$id(A) = A$$ is compact, and since $$(B,\tau_d)$$ is Hausdorff then $$A$$ is closed. Hence the image of every closed set under the identity is closed.

This finishes the proof that $$d$$ defines the same topology.

• Now this is not true in $$M$$. What is happening is that $$d$$ is not $$w^*$$ continuous. [Nick proposed another proof for this part that seems more efficient than this] Let's see: If $$d$$ defines the same topology on $$M$$ then $$B_d(0,1)$$ is $$w*$$-open. Since $$X = C_0([0,1])$$ has $$dim = \infty$$, then every neighborhood of zero in the weak star topology of $$X'$$ contains a subspace of infinite dimension, in particular $$B_d(0,1)$$ is not bounded and this is absurd. (In the comments of Nick's answer I give an argument more complicated using a linear functional $$H$$)